ELECTRONIC
COMMUNICATIONS in PROBABILITY
INFINITE DIVISIBILITY OF GAUSSIAN SQUARES WITH NON–ZERO MEANS
MICHAEL B. MARCUS1
The City College of New York, CUNY email: [email protected] JAY ROSEN2
The City College of New York, CUNY email: [email protected]
Submitted January 2, 2008, accepted in final form June 9, 2008 AMS 2000 Subject classification: 60G15, 60E07, 60J27
Keywords: Gaussian vectors, infinite divisibility, Markov chains Abstract
Letη = (η1, . . . , ηn) be anRn valued Gaussian random variable andc= (c1, . . . , cn) a vector inRn. We give necessary and sufficient conditions for ((η1+c1α)2, . . . ,(ηn+ cnα)2) to be infinitely divisible for allα∈R1, and point out how this result is related to local times of Markov chains determined by the covariance matrix of η.
1 Introduction
Let η = (η1, . . . , ηn) be an Rn valued Gaussian random variable. η is said to have infinitely divisible squares ifη2:= (η12, . . . , ηn2) is infinitely divisible, i.e. for anyrwe can find anRn valued random vectorZrsuch that
η2law= Zr,1+· · ·+Zr,r, (1.1) where {Zr,j}, j = 1, . . . , r are independent identically distributed copies of Zr. We express this by saying thatη2 is infinitely divisible.
We are interested in characterizing Gaussian processes with infinitely divisible squares which do not have mean zero. We setηi=Gi+ci,EGi= 0,i= 1, , . . . , n. Let Γ be the covariance matrix of (G1, . . . , Gn) and set
c:= (c1, . . . , cn). (1.2) Set
G+c:= (G1+c1, . . . , Gn+cn) (1.3)
1RESEARCH SUPPORTED BY NSF AND PSCCUNY
2RESEARCH SUPPORTED BY NSF AND PSCCUNY
364
and
(G+c)2:= ((G1+c1)2, . . . ,(Gn+cn)2). (1.4) In order to continue we need to define different types of matrices. LetA={ai,j}1≤i,j≤n be ann×nmatrix. We callA a positive matrix and writeA≥0 ifai,j ≥0 for all i, j. We write A >0 ifai,j >0 for all i, j. We say that Ahas positive row sums if Pn
j=1ai,j≥0 for all 1≤i≤n.
The matrixAis said to be an M matrix if (1) ai,j≤0 for alli6=j.
(2) A is nonsingular andA−1≥0.
This definition includes the trivial 1×1 M matrix witha1,1>0.
A matrix is called a signature matrix if its off diagonal entries are all zero and its diagonal entries are either one or minus one.
Our starting point is a theorem by Griffiths and Bapat [1, 7], (see also [8, Theorem 13.2.1]) that characterizes mean zero Gaussian vectors with infinitely divisible squares.
Theorem 1.1(Griffiths, Bapat).LetG= (G1, . . . , Gn)be a mean zero Gaussian ran- dom variable with strictly positive definite covariance matrixΓ ={Γi,j}={E(GiGj)}. Then G2 is infinitely divisible if and only if there exists a signature matrix N such that
NΓ−1N is an M matrix. (1.5)
The role of the signature matrix is easy to understand. It simply accounts for the fact that ifGhas an infinitely divisible square, then so does (ǫ1G1, . . . , ǫnGn) for any choice ofǫi =±1,i= 1, . . . , n. Therefore, if (1.5) holds forN with diagonal elements n1, . . . , nn
¡NΓ−1N¢−1
=NΓN ≥0 (1.6)
since the inverse of an M matrix is positive. Thus (n1G1, . . . , nnGn) has a positive covariance matrix and its inverse is anM matrix.. (For this reason, in studying mean zero Gaussian vectors with infinitely divisible squares one can restrict ones attention to vectors with positive covariance.)
The question of characterizing Gaussian random variables with non-zero mean and infinitely divisible squares first came up in the work of N. Eisenbaum [2, 3] and then in joint work by Eisenbaum and H. Kaspi [4], in which they characterize Gaussian processes with a covariance that is the 0-potential density of a symmetric Markov process. This work is presented and expanded in [8, Chapter 13]. The following theorem is taken from Theorem 13.3.1 and Lemma 13.3.2 in [8].
Theorem 1.2 (Eisenbaum, Kaspi). Let G = (G1, . . . , Gn) be a mean zero Gaus- sian random variable with strictly positive definite covariance matrix Γ = {Γi,j} = {E(GiGj)}. Let1= (1, . . . ,1)∈Rn. Then the following are equivalent:
(1) G+1αhas infinitely divisible squares for allα∈R1;
(2) Forξ=N(0, b2)independent ofG,(G1+ξ, . . . , Gn+ξ, ξ)has infinitely divisible squares for some b6= 0. Furthermore, if this holds for some b6= 0, it holds for all b∈R1, withN(0,0) = 0.
(3) Γ−1 is an M matrix with positive row sums.
To avoid having to comment about trivial exceptions to general statements we assume that Gcan not be written as the direct sum of two independent Gaussian vectorsG′ and G′′. This is equivalent to saying that the covariance matrix of G is irreducible.
We have the following description of Gaussian vectors with infinitely divisible squares that doesn’t require that each component of the vector has the same mean.
Theorem 1.3. Let G= (G1, . . . , Gn)be a mean zero Gaussian random variable with irreducible strictly positive definite covariance matrix Γ ={Γi,j}={E(GiGj)}. Let c= (c1, . . . , cn)∈Rn,c6=0and letC be a diagonal matrix with ci=Ci,i,1≤i≤n . Then the following are equivalent:
(1) G+cαhas infinitely divisible squares for allα∈R1;
(2) For ξ = N(0, b2) independent of G, (G1+c1ξ, . . . , Gn+cnξ, ξ) has infinitely divisible squares for someb 6= 0. Furthermore, if this holds for someb 6= 0, it holds for allb∈R1;
(3) CΓ−1C is anM matrix with positive row sums.
We list several consequences of Theorem 1.3. An important step in the proof of Theorem 1.3 is to show that, under the hypotheses of this Theorem, no component of c can be equal to zero. We state this as 1. of the next Corollary, and explore its implications.
Corollary 1.1. LetG= (G1, . . . , Gn)be a mean zero Gaussian random variable with irreducible strictly positive definite covariance matrix Γ ={Γi,j}={E(GiGj)}. Let c= (c1, . . . , cn)∈Rn,c6=0and letC be a diagonal matrix with ci=Ci,i,1≤i≤n . Then
1. When any of the equivalent conditions (1), (2) and (3) of Theorem 1.3 hold no component ofc can be equal to zero;
2. IfG2 is infinitely divisible none of the entries of Γare equal to zero;
3. When any of the equivalent conditions (1), (2) and (3) of Theorem 1.3 hold and Γ≥0, (in which case, by 2., Γ>0), thencicj>0,1≤i, j≤n;
4. When CΓ−1C is an M matrix, it follows that NΓ−1N is also an M matrix, where the diagonal elements of the signature matrixN areni=signci,1≤i≤ n.
To elaborate on Corollary 1.1,4. we know by (3)⇒(1) of Theorem 1.3 and Theorem 1.1 that when CΓ−1C is anM matrix there exists a signature matrix N such that NΓ−1N is anM matrix. In Corollary 1.1,4. we show howCand N are related.
In the next corollary we use Theorem 1.3 to obtain two properties of the Gaussian vectorGthat hold whenG2 is infinitely divisible.
Corollary 1.2. LetG= (G1, . . . , Gn)be a mean zero Gaussian random variable with irreducible strictly positive definite covariance matrix Γ ={Γi,j} = {E(GiGj)} and suppose thatGhas infinitely divisible squares. Set
hj,n= Γj,n
Γn,n
1≤j≤n−1; (1.7)
1. Then
(G1+h1,nα, . . . , Gn−1+hn−1,nα, Gn+α) (1.8) has infinitely divisible squares for allα∈R1.
2. Write
G= (η1+h1,nGn, . . . , ηn−1+hn−1,nGn, Gn) (1.9) where
ηj =Gj−hj,nGn 1≤j≤n−1. (1.10) Then
(η1+h1,nα, . . . , ηn−1+hn−1,nα) (1.11) has infinitely divisible squares for allα∈R1.
3. Let E denote the covariance matrix of (η1, . . . ηn−1). If Γ ≥ 0, E−1 is an M matrix.
Remark 1.1. In particular Corollary 1.2, 1. shows that when a Gaussian vectorGin Rn has infinitely divisible squares there exists a vectorc= (c1, . . . , cn) for whichG+c has infinitely divisible squares. Also Corollary 1.2, 2. shows that when a Gaussian vector G in Rn has infinitely divisible squares, then (η1, . . . , ηn−1), the orthogonal complement of the projection ofGontoGn, has infinitely divisible squares.
We next list several properties of the elements of the covariance matrix Γ ={Γi,j}= {E(GiGj)} that hold when G2 is infinitely divisible, or when any of the equivalent conditions (1), (2) and (3) of Theorem 1.2 hold. The first two are known, references for them are given in Remark 2.1.
Corollary 1.3. LetG= (G1, . . . , Gn)be a mean zero Gaussian random variable with irreducible strictly positive definite covariance matrix Γ = {Γi,j} ={E(GiGj)}. Let c= (c1, . . . , cn)∈Rn,c6=0. Then
1. WhenG2 is infinitely divisible, Γ≥0 andn≥3
Γj,lΓk,l≤Γj,kΓl,l ∀1≤j, k, l≤n. (1.12) 2. When any of the equivalent conditions (1), (2) and (3) of Theorem 1.2 hold,
0<Γi,j≤Γi,i∧Γj,j. (1.13) 3. When any of the equivalent conditions (1), (2) and (3) of Theorem 1.3 hold
Γi,i≥ ci
cj
Γi,j ∀1≤i, j≤n; (1.14)
4. Whenn= 2and the covariance matrix ofGis invertible,(G+cα)2is infinitely divisible for allα∈R1 if and only if
Γi,i≥ ci
cj
Γi,j>0 ∀1≤i, j≤2. (1.15)
5. When n ≥ 2 there is no Gaussian vector G for which (G+cα)2 is infinitely divisible for all α∈R1 and allc ∈Rn. (Recall that we are assuming that the covariance matrix ofG is irreducible. This rules out the possibility that all the components ofGare independent.)
By definition, when (G+c)2 is infinitely divisible, it can be written as in (1.1) as a sum of r independent identically distributed random variables, for allr≥1. Based on the work of Eisenbaum and Kaspi mentioned above and the joint paper [5] we can actually describe the decomposition. (In fact this decomposition plays a fundamental role in the proofs of Lemma 4.1 and Theorem 4.2, [2] and Theorem 1.1, [3].) We give a rough description here. For details see [2, 3, 4] and [8, Chapter 13].
Assume that (1), (2) and (3) of Theorem 1.3 hold. ConsiderG/c, (see (2.1)). Let Γc
denote the covariance matrix ofG/c. Theorem 1.2 holds forG/cand Γc, so Γ−1c is an M matrix with positive row sums. To be specific letG/c∈Rn. SetS ={1, . . . , n}. By [8, Theorem 13.1.2] Γc is the 0-potential density of a strongly symmetric transient Borel right process, say X, onS. We show in the proof of [8, Theorem 13.3.1] that we can find a strongly symmetric recurrent Borel right process Y on S∪ {0} with Px(T0 <∞)>0 for allx∈S such thatX is the process obtained by killingY the first time it hits 0. LetLxt ={Lxt;t∈R+, x∈S∪ {0}}denote the local time ofY. It follows from the generalized second Ray-Knight Theorem in [5], see also [8, Theorem 8.2.2] that underP0×PG,
½
Lxτ(t)+1 2
µGx
cx
¶2
; x∈S
¾law
=
½1 2
µGx
cx
+√ 2t
¶2
;x∈S
¾
(1.16) for allt∈R+, whereτ(t) = inf{s >0|L0s> t}, the inverse local time at zero, andY andGare independent. Consequently
nc2xLxτ(α2/2)+1
2G2x;x∈Solaw
= n1
2(Gx+cxα)2;x∈So
(1.17) for allα∈R1. (We can extendαfromR+toR1becauseGis symmetric.) {c2xLxτ(α2/2); x∈ S} and {12G2x;x∈S} are independent. G2 is infinitely divisible and for all integers r≥1
c2·L·τ(α2/2)
law= c2·L·τ(α2/(2r)),1+· · ·+c2·L·τ(α2/(2r)),r (1.18) where {L·τ(α2/(2r)),j}, j= 1, . . . , rare independent.
Note that in (1.17) we identify the components of the decomposition of{(Gx+cxα)2;x∈ S} that mark it as infinitely divisible. The profound connection between Gaus- sian processes with infinitely divisible squares and local times of strongly symmet- ric Markov processes shows that the question of Gaussian processes with infinitely divisible squares is more than a mere curiosity.
In this same vein we can describe a decomposition of mean zero Gaussian vectors with infinitely divisible squares that is analogous to (1.17). Suppose G2 is infinitely divisible. WriteGandη1, . . . , ηn−1 as in (1.9) and (1.10) and set
ci = Γi,n
Γn,n
, i= 1, . . . , n. (1.19)
LetP := (η1, . . . ηn−1) andc:=c1, . . . , cn−1.
It follows from Corollary 1.2, 2, that η+cα has infinitely divisible squares for all α∈R1. Therefore, as in (1.17),
nc2xLxτ(α2/2)+1
2ηx2;x∈Solaw
= n1
2(ηx+cxα)2;x∈So
(1.20) for allα∈R1. HereLxt is the local time determined by the processP/c, in the same way as the local time is determined byG/cin the paragraph containing (1.16). Letξ be a normal random variable with mean zero and varianceEG2n/2 that is independent of everything else in (1.20). It follows from (1.20) and (1.9) that
nc2xLxτ(ξ2)+1
2ηx2;x∈Solaw
= n1
2G2x;x∈So
. (1.21)
This isomorphism identifies the components of the decomposition of{G2x;x∈S}that mark it as infinitely divisible.
There remains an interesting question. Assume thatG2has infinitely divisible squares.
Is it possible for (G1+α, . . . , Gn+α) to have infinitely divisible squares for some α6= 0 but not for allα∈R1? We do know that if this is the case there would exist an 0< α0<∞such that (G1+α, . . . , Gn+α) would have infinitely divisible squares for all 0≤ |α| ≤α0 but not for|α|> α0. We can show that such an α0 always exists for (G1+α, G2+α), as long asEG1G26= 0, [9].
2 Proof of Theorem 1.3
Proof of Theorem 1.3 If none of the ci, i = 1, . . . , n are equal to 0, the theorem follows directly from Theorem 1.2 applied to the Gaussian process
G c =
µG1
c1, . . . ,Gn
cn
¶
(2.1) which is a mean zero Gaussian random variable with strictly positive definite covari- ance matrix
Γc=
½Γi,j
cicj
¾
=
½ E
µGi
ci
Gj
cj
¶ ¾
=C−1ΓC−1, (2.2)
so that Γ−1c =CΓ−1C. Applying Theorem 1.2 gives Theorem 1.3 except that (1) and (2) are replaced by
(1′) µ
G1
c1
+α, . . . ,Gn
cn
+α
¶
has infinitely divisible squares. (2.3)
(2′) µ G1
c1
+ξ, . . . ,Gn
cn
+ξ, ξ
¶
has infinitely divisible squares. (2.4) It is easy to see that these hold if and only if (1) and (2) hold.
Therefore, to complete the proof of Theorem 1.3 we need only show that each of the conditions (1), (2), and (3) imply that none of theci, i= 1, . . . , nare equal to 0. This is obvious for (3) since an M matrix is invertible.
To proceed we give a modification of [8, Lemma 13.3.2].
Lemma 2.1. LetG= (G1, . . . , Gn)be a mean zero Gaussian process with covariance matrix Γ which is invertible. Consider G = (G1+c1ξ, . . . , Gn+cnξ, ξ) where ξ = N(0, b2),b6= 0, is independent ofG. Denote the covariance matrix ofGby Γ. Then
Γj,k = Γj,k j, k= 1, . . . , n
Γn+1,k = −Pn
j=1cjΓj,k k= 1, . . . , n Γn+1,n+1 = 1
b2+ Xn j,k=1
cjckΓj,k
(2.5)
where, for an invertible matrix Awe use Ai,j to denote {A−1}i,j.
Note that it is possible for some or all of the components ofc to be equal to zero.
Proof To prove (2.5) we simply go through the elementary steps of taking the inverse of Γ. We begin with the array
Γ1,1+c21b2 . . . Γ1,n+c1cnb2 c1b2 ¯¯ 1 . . . 0 0
... . .. ... ...
¯¯
¯¯ ... . .. ... ... Γn,1+cnc1b2 . . . Γn,n+b2 cnb2 ¯¯ 0 . . . 1 0 c1b2 . . . cnb2 b2 ¯¯ 0 . . . 0 1
(2.6)
Next, for eachj= 1, . . . , nsubtractcj times the last row from thej–th row and then divide the last row byb2to get
Γ1,1 . . . Γ1,n 0 ¯¯ 1 . . . 0 −c1
... . .. ... ...
¯¯
¯¯ ... . .. ... ... Γn,1 . . . Γn,n 0 ¯¯ 0 . . . 1 −cn
c1 . . . cn 1 ¯¯ 0 . . . 0 1/b2
(2.7)
This shows that det(Γ) = b2det(Γ) and consequently Γ is invertible if and only if Γ is invertible.
We now work with the firstnrows to get the inverse of Γ so that the array looks like 1 . . . 0 0 ¯¯ Γ1,1 . . . Γ1,n a1
... . .. ... ...
¯¯
¯¯ ... . .. ... ... 0 . . . 1 0 ¯¯ Γn,1 . . . Γn,n an
c1 . . . cn 1 ¯¯ 0 . . . 0 1/b2
(2.8)
At this stage we don’t know what are theaj,j= 1, . . . , n.
Finally, for eachj= 1, . . . , nwe subtractcj times thej–th row from the last row to obtain
1 . . . 0 0 ¯¯ Γ1,1 . . . Γ1,n a1
... . .. ... ...
¯¯
¯¯ ... . .. ... ... 0 . . . 1 0 ¯¯ Γn,1 . . . Γn,n an
0 . . . 0 1 ¯¯ −Pn
j=1cjΓj,1 . . . −Pn
j=1cjΓj,n an+1
(2.9)
where
an+1= (1/b2− Xn k=1
ckak). (2.10)
Since the inverse matrix is symmetric we see thatak =−Pn
j=1cjΓj,k, k= 1, . . . , n.
This verifies (2.5).
We now show that (2) implies that no component of c can be zero. Recall that c 6= 0.ˆEˆE Suppose that some of the components of c are equal to zero. We can separate the components ofG1, . . . , Gn into two sets. Those that have the associated ci 6= 0, 1≤i≤nand those that have the associatedcj = 0, 1≤j ≤n. There must be at least one member of the first set that is not independent of some member of the second set. Otherwise Γ is not irreducible. (Recall that this is an hypothesis, stated just before Theorem 1.3.) We take these two members and relabel themG1 andG2
. If (2) ˆEholds then Ge = (G1+c1ξ, G2, ξ) has an infinitely divisible square. LetˆE Γ denote the covariance matrix ofe G. Then by Theorem 1.1 there exists a signaturee matrixN such that thatNeΓ−1N is anM matrix.ˆE It follows from (2.9) that
NeΓ−1N =
Γ1,1(2) Γ1,2(2)n1n2 −c1Γ1,1(2)n1n3
Γ2,1(2)n1n2 Γ2,2(2) −c1Γ1,2(2)n2n3
−c1Γ1,1(2)n1n3 −c1Γ1,2(2)n2n3 1/b2+c21Γ1,1(2)
. (2.11)
where Γ(2) is the covariance matrix ofˆE (G1, G2) and Γi,j(2) = (Γ−1(2))i,j. SinceNeΓ−1N is anM matrix, it must have negative off diagonal elements. In fact they are strictly negative becauseG1 andG2 are not independent. Therefore
0<(NΓe−1N)1,3(NeΓ−1N)2,3=c21Γ1,12 Γ1,22 n1n2
which implies that (NeΓ−1N)1,2 = Γ1,22 n1n2 > 0.ˆE This contradiction shows that when (2) holds no component ofc can be zero.
We complete the proof of the Theorem by showing that (1)⇒(2) , which, in partic- ular, implies that when (1) holds no component ofccan be zero.
It follows from (1) that (√mα, G1+√mαc1, . . . , Gn+√mαcn) has infinitely divisible squares for all integers m ≥ 0 and α ∈ R1. Let G0 := 0, c0 := 1 and let λ = (λ1, . . . , λn) be an n-dimensional vector and Λ ann×ndiagonal matrix with λj as
itsj-th diagonal entry. By [8, Lemma 5.2.1]
Eexp Ã
− Xn i=0
λi(Gi+√mαci)2/2
!
(2.12)
= 1
(det(I+ Γ Λ))1/2exp Ã
m α2 Ã
−λ0
2 −cΛct
2 +(cΛeΓ Λct) 2
!!
, where
eΓ := (Γ−1−Λ)−1= (I−Γ Λ)−1Γ. (2.13) By the same lemma
Eexp Ã
− Xn i=0
λi(Gi+ciξ)2/2
!
(2.14)
= 1
(det(I+ Γ Λ))1/2exp Ã
ξ2 Ã
−λ0
2 −cΛct
2 +(cΛeΓ Λct) 2
!!
.
Compare (2.12) and (2.14) with (13.82) and (13.83) in the proof of (2) ⇒(3) of [8, Theorem 13.3.1]. Following that proof we see that (1)⇒(2).
Proof of Corollary 1.1
1. As observed in the proof of Theorem 1.3, (3), CΓ−1C is invertible and thus detC6= 0.
2. ˆE Pick any component ofG; for convenience we takeGn. LetˆE ηj =Gj− Γj,n
Γn,n
Gn 1≤j≤n−1. (2.15)
We write
G= µ
η1+Γ1,n
Γn,nGn, . . . , ηn−1+Γn−1,n
Γn,n Gn, Gn
¶
(2.16) and note that this has the form of
G= (η1+c1ξ, . . . , ηn−1+cn−1ξ, ξ) (2.17) whereξ=N(0, EG2n) is independent of η= (η1, . . . , ηn−1).
If the covariance matrix of (η1, . . . , ηn−1) is irreducible then the fact thatG2is infinitely divisible and 1. of this corollary imply that cj 6= 0, 1 ≤j ≤n−1.
Thus none of Γj,n, 1≤j ≤n−1, are equal to zero. Since the initial choice of Gn is arbitraryˆE we get 2.
Now suppose that the covariance matrix of (η1, . . . , ηn−1) is not irreducible. As- sume that Γ1,n= 0. Ifη1 is independent of (η2, . . . , ηn−1), then sinceη1 =G1, and it is independent of Gn, we get a contradiction of the hypothesis that the covariance matrix ofGis irreducible. Therefore let (η1, . . . , ηl), 2≤l≤n−2, be the smallest set of components in (η1, . . . , ηn−1) that both containsη1and has
an irreducible covariance matrix. Consider the corresponding Gaussian vector (G1, . . . , Gl), Note that at least one of the components in this vector is not in- dependent ofGn, since if this is not the case (G1, . . . , Gl) would be independent of (Gl+1, . . . , Gn) which contradicts the hypothesis that the covariance matrix ofGis irreducible.
SupposeG2 depends onGn, and consider the vector (G1, G2, Gn). This vector has infinitely divisible squares because G has infinitely divisible squares. Let Gn := ξ. Then we can write (G1, G2, Gn) as (η1, η2+αξ, ξ), where α 6= 0.
However, by Theorem 1.3, or 1. of this corollary, this vector does not have infinitely divisible squares. This contradiction shows that none of Γj,n, 1 ≤ j ≤ n−1, are equal to zero, and as above this establishes 2. even when the covariance matrix of (η1, . . . , ηn−1) is not irreducible.
3. By Theorem 1.3, (3)
¡CΓ−1C¢−1
j,k= Γj,k
cjck ≥0. (2.18)
and by hypothesis and 2., Γj,k >0. (Also by 1. neithercj norck are equal to zero.) Therefore under the hypotheses of 3. we actually have
¡CΓ−1C¢−1 j,k= Γj,k
cjck
>0, (2.19)
so, since Γj,k>0,cjck >0.
4. WriteC=CN N forN as given. Then consider
CN NΓ−1NCN =CΓ−1C. (2.20) SinceCN >0 we see thatNΓ−1N is anM matrix.
Proof of Corollary 1.2
1. To begin suppose that Γ ≥ 0. In this case we see by Corollary 1.1, 2. that hj,n>0, 1≤j ≤n−1. SinceGhas infinitely divisible squares so does
G h =
µ G1
h1,n
, . . . , Gn−1
hn−1,n
, Gn
¶
. (2.21)
WriteGas in (2.16) so that G h =
µ η1
h1,n
+Gn, . . . , ηn−1
hn−1,n
+Gn, Gn
¶
. (2.22)
Let Θ be the covariance matrix ofG/h. Since Γ≥0 andhj,n>0, 1≤j≤n−1, we have Θ ≥ 0. Therefore by Theorem 1.1, Θ−1 is an M-matrix. It is also a matrix of the type Γ−1 considered in [8, Lemma 13.4.1]. It follows from (4) =⇒ (3) of this Lemma that Θ−1 has positive row sums. Therefore by Theorem 1.2
µ η1
h1,n
+Gn+α, . . . , ηn−1
hn−1,n
+Gn+α, Gn+α
¶
(2.23)
has infinitely divisible squares for allα∈R1. This is equivalent to (1.8).
In general letN be the signature matrix such thatNΓN ≥0. Without loss of generality we can takenn= 1. Applying the result in the preceding paragraph to (n1G1, . . . , nn−1
Gn−1, Gn) we see that
(n1G1+n1h1,nα, . . . , nn−1Gn−1+nn−1hn−1,nα, Gn+α) (2.24) has infinitely divisible squares for allα∈R1. This implies that
(G1+h1,nα, . . . , Gn−1+hn−1,nα, Gn+α) (2.25) has infinitely divisible squares, which gives (1.8).
2. ˆEWriteGas in (1.9). If the covariance matrix ofη= (η1, . . . , ηn−1) is irreducible then it follows from Theorem 1.3 and the hypothesis thatG2 is infinitely divis- ible, that (η+hα)2 is infinitely divisible for all α. However, it is easy to see that the covariance matrix of (η1, . . . , ηn−1) need not be irreducible. (Consider G= (η1+ξ, η2+ξ, ξ) whereη1,η2,ξare i.i.d. N(0,1). G2is infinitely divisible but, obviously, the covariance matrix of (η1, η2) is not irreducible.)
If the covariance matrix of (η1, . . . , ηn−1) is not irreducible, consider a subset of the components say (η1, . . . , ηl) such that its covariance matrix is irreducible.
SinceG2 is infinitely divisible
G= (η1+h1,nGn, . . . , ηl+hl,nGn, Gn) (2.26) is infinitely divisible. It follows from the argument in the beginning of the preceding paragraph that (η1+h1α, . . . , ηl+hlα,) has infinitely divisible squares for allα. This holds for all the blocks of irreducible components ofη. Therefore (η+hα)2 is infinitely divisible for allα.
3. As we point out in 2., immediately above, the covariance matrix of (η1, . . . , ηn−1) is not necessarily irreducible. If it is not, as above, consider a subset of the components, say (η1, . . . , ηl), such that its covariance matrix is irreducible. Since G2 is infinitely divisible
G= (η1+h1,nGn, . . . , ηl+hl,nGn, Gn) (2.27) is infinitely divisible.ˆE LetE′ be the covariance matrix of (η1, . . . , ηl). It follows from Theorem 1.3, thatC′E′−1C′is anM matrix, whereC′is a diagonal matrix with entries ci = hi,n, 1 ≤ i ≤ l. Since Γ ≥ 0, the diagonal elements of C′ are strictly positive. ThusE′−1 is an M matrix. It is easy to see that if the covariance matrix of each irreducible block of (η1, . . . , ηn−1) is an M matrix, thenE is anM matrix.
Proof of Corollary 1.3
1. LetE be as in Corollary 1.2, 3. SinceE ≥0 we see that Ej,k=Eηjηk = Γj,k−Γj,nΓk,n
Γn,n ≥0. (2.28)
This gives (1.12) when l =n. Since the choice of Gn in 8. is arbitrary we get (1.12) as stated.
3. By (2.19)
Γj,k
cjck
>0 1≤j, k≤n. (2.29)
Consider (Gi+ci, Gj+cj),i6=j. Let Γ(2) be the covariance matrix of (Gi, Gj) and Γc be the covariance matrix of (Gi/ci, Gj/cj) so that Γc = Ce−1Γ(2)Ce−1 whereCe is a diagonal matrix with entriesci, cj. Hence Γ−1c =CΓe −1(2)C. Consid-e ering the nature of the inverse of the 2×2 matrix Γ(2) we have that
Γi,ic =c2iΓi,i=c2iΓj,j
det Γ >0 i6=j (2.30) and
Γi,jc =cicjΓi,j=−cicjΓi,j
det Γ . (2.31)
Since (Gi+ci, Gj+cj) has infinitely divisible squaresCΓ−1Chas positive row sums. Therefore
c2iΓj,j ≥cicjΓi,j (2.32) which gives (1.14).
4. If (1.15) holds no component of ccan be equal to zero and Γi,j6= 0. The proof of 3. shows that when n = 2 the first inequality in (1.15) is necessary and sufficient forCΓ−1Cto have positive row sums. The second inequality in (1.15) is necessary and sufficient forCΓ−1C to be an M-matrix.
2. This follows from 3.
5. This follows from 3. and Corollary 1.1, 1.
Remark 2.1. Item 1. in Corollary 1.3 is given in [8, (13.39)]. An interesting con- sequence of this property is given in [8, (13.42)]. Item 1. in Corollary 1.2 follows from [8, Theorem 13.3.3 and Theorem 13.3.1]. Item 2. in Corollary 1.3 follows from [8, Theorem 13.3.1 and (13.2)]. These are all are consequences of the relationship betweenM matrices with positive row sums and the 0-potential density of Borel right process. Item 5. in Corollary 1.3 is also a consequence of [3], Remark 2.3.
Example 2.1. LetG= (G1, G2) have mean zero and have covariance matrix Γ =
µ 1 43
4
3 2
¶
, Γ−1=
µ 9 −6
−6 9/2
¶
. (2.33)
By Corollary 1.3, 4. (G1+α, G2+α) does not have infinitely divisible squares for all α∈R1, and obviously, Γ−1 does not have positive row sums. However, by Corollary 1.3, 4. again or Corollary 1.2, 1., (G1+α, G2+ (4/3)α) does have infinitely divisible squares for allα∈R1. (LetC be the diagonal matrix withC1,1= 1 andC2,2= 4/3.
Then
CΓ−1C=
µ 9 −8
−8 8
¶
(2.34) is anM matrix with positive row sums.)
Moreover by Corollary 1.3, 4. (G1+α, G2+c2α) has infinitely divisible squares for allα∈R1, for all 1≤c2≤2.
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